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JEE Main 2016 · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2016

Q110JEE Main 2016Difficult

Electric Flux & Gauss's Law

The region between two concentric spheres of radii aa and b,b, respectively (see figure), has volume charge density ρ=Ar,\rho = \frac{A}{r}, where AA is a constant and rr is the distance from the center. At the center of the spheres is a point charge QQ. The value of AA such that the electric field in the region between the spheres will be constant, is:

Figure

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Answer: (c)

  1. Charge within rr (a<r<ba < r < b): q=Q+∫arAr4πr2dr=Q+2πA(r2−a2)q = Q + \int_a^r \frac{A}{r}4\pi r^2 dr = Q + 2\pi A(r^2 - a^2).
  2. E=q4πε0r2=14πε0(2πA+Q−2πAa2r2)E = \frac{q}{4\pi\varepsilon_0 r^2} = \frac{1}{4\pi\varepsilon_0}\left(2\pi A + \frac{Q - 2\pi A a^2}{r^2}\right)
  3. For EE to be constant the 1r2\frac{1}{r^2} term must vanish: Q=2πAa2⇒A=Q2πa2Q = 2\pi A a^2 \Rightarrow A = \frac{Q}{2\pi a^2}.
  4. So the answer is (c).
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